Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $z = \dfrac{y^2 + 10y}{y^2 + 8y - 20} \times \dfrac{y - 2}{y + 6} $
Explanation: First factor the quadratic. $z = \dfrac{y^2 + 10y}{(y - 2)(y + 10)} \times \dfrac{y - 2}{y + 6} $ Then factor out any other terms. $z = \dfrac{y(y + 10)}{(y - 2)(y + 10)} \times \dfrac{y - 2}{y + 6} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ y(y + 10) \times (y - 2) } { (y - 2)(y + 10) \times (y + 6) } $ $z = \dfrac{ y(y + 10)(y - 2)}{ (y - 2)(y + 10)(y + 6)} $ Notice that $(y + 10)$ and $(y - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ y(y + 10)\cancel{(y - 2)}}{ \cancel{(y - 2)}(y + 10)(y + 6)} $ We are dividing by $y - 2$ , so $y - 2 \neq 0$ Therefore, $y \neq 2$ $z = \dfrac{ y\cancel{(y + 10)}\cancel{(y - 2)}}{ \cancel{(y - 2)}\cancel{(y + 10)}(y + 6)} $ We are dividing by $y + 10$ , so $y + 10 \neq 0$ Therefore, $y \neq -10$ $z = \dfrac{y}{y + 6} ; \space y \neq 2 ; \space y \neq -10 $